Problem: Solve for $x$ : $2\sqrt{x} + 9 = 6\sqrt{x} + 3$
Answer: Subtract $2\sqrt{x}$ from both sides: $(2\sqrt{x} + 9) - 2\sqrt{x} = (6\sqrt{x} + 3) - 2\sqrt{x}$ $9 = 4\sqrt{x} + 3$ Subtract $3$ from both sides: $9 - 3 = (4\sqrt{x} + 3) - 3$ $6 = 4\sqrt{x}$ Divide both sides by $4$ $\frac{6}{4} = \frac{4\sqrt{x}}{4}$ Simplify. $\dfrac{3}{2} = \sqrt{x}$ Square both sides. $\dfrac{3}{2} \cdot \dfrac{3}{2} = \sqrt{x} \cdot \sqrt{x}$ $x = \dfrac{9}{4}$